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how to prove n sigma r=1=n(n+1)/2
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\[\sum_{r=1}^{n}r=n(n+1)/2\]
how to prove this
you may find this interesting :) http://mathcentral.uregina.ca/QQ/database/QQ.02.06/jo1.html
but how to prove it
First use: \[ \sum r = 1+2+\dots +n = n + (n-1) + \dots +1 = \sum (n+1)-r \]
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Therefore \[ 2\sum r = \sum r + \sum(n+1)-r = \sum r + (n+1) - r = \sum n+1 = n(n+1) \]So \[ 2\sum r = n(n+1) \implies \sum r = \frac{n(n+1)}{2} \]
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